A number $x$ is called normal in base $b$ if every sequence of base $b$ digits $b_1b_2...b_n$ occurs with natural density $1/b^n$ in the decimal expansion of $x$.

There exist numbers normal in every base (called absolutely normal) and irrational numbers normal in no base (called absolutely non-normal), an example is given here.

Is it known whether there exist numbers that are normal in every base except one or numbers non-normal in every base except one?

The question can be stated rather easily but an answer probably will take a lot of effort, so thanks in advance, also for any reference to literature :).

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    $\begingroup$ I first read this as "every base except for base 1". $\endgroup$
    – asmeurer
    Sep 25, 2016 at 17:33
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    $\begingroup$ A bit more general question comes to one's mind: How small can be $S$ such that there exists $x\in\mathbb{R}$ that is normal in all bases $b\in\mathbb{N}\setminus S$? Is it possible that we can take $S=\{2,4,8,16,\dots\}$? $\endgroup$
    – yo'
    Sep 25, 2016 at 18:04
  • $\begingroup$ @yo' Yes, numbers exist that are as "selectively" normal as you want, subject only to the exercise/theorem in Watson's answer. $\endgroup$ Sep 26, 2016 at 2:53
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    $\begingroup$ @yo' - I think the question you are asking is the "real" question here. The observation that "base 2 normal number" and "base 4 normal number" is intuitively obvious - we are looking at the same expansion, except in one case we are grouping the digits in twos. To properly write that down in full generality one surely needs to do some work, but the outcome is hardly surprising. The tricky bit is comparing, say, base 2 normality with base 3 normality. $\endgroup$ Sep 26, 2016 at 10:10

3 Answers 3


This is not possible. In Kuipers, L. and Niederreiter, H. Uniform Distribution of Sequences (1974), you can find the following exercise, on page $77$:

If $b_1$ and $b_2$ are integers $≥2$, such that one is a rational power of the other, then $a$ is normal to the base $b_1$ if and only if $a$ is normal to the base $b_2$.

(This is also theorem 2.7. in Bailey, D. H. and Crandall, R. E. On the Random Character of Fundamental Constant Expansions, Experiment. Math., Volume 10, Issue 2 (2001), 175-190.)

Therefore, if $x$ is a real number that is normal in every base but one, say $b_1$, then actually it wouldn't be normal to the base $b_1^2 \neq b_1$, contradicting the assumption. The situation is similar for numbers that are non-normal in every base except one.


Theorem 1

If $x$ is normal in infinitely many bases of the powers of $b^k$ then it is normal in base $b$.

Conclusion: This rules out your first curiosity.

Theorem 2:

If $x$ is not normal in base $b^k$ then it cannot be normal in base $b$.

Conclusion: This rules out your second curiosity.

Source for the theorems: On normal numbers, Veronica Becher, page 20-21.

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    $\begingroup$ Thanks, I voted on your answer, but the other answer was given 5 minutes earlier so I will accept that one :) $\endgroup$
    – s.harp
    Sep 25, 2016 at 12:25
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    $\begingroup$ @s.harp : You should accept the answer that you feel best answers your question. Maybe speediness is part of your assessment, maybe not. $\endgroup$ Sep 25, 2016 at 16:24
  • $\begingroup$ @EricTowers in this case I think the answers are too similar for me to have much of a preference. $\endgroup$
    – s.harp
    Sep 25, 2016 at 21:24

The most general answer to date for your question comes from a paper of Schmidt (http://matwbn.icm.edu.pl/ksiazki/aa/aa7/aa7311.pdf). Schmidt showed that if $A\subset \mathbb{N}_{\ge 2}$ is a set closed under multiplicative dependence*, then there exist uncountably many numbers that are normal to every base $b\in A$ and not normal to every base $b\not \in A$. And by earlier work of Maxfield and Cassels, if $A$ is not closed under multiplicative dependence, then no such number exists. Since if one element is missing from (resp., included in) a set closed under multiplicative dependence, infinitely many are missing (resp, included). Thus, the answer to your question is no.

*A set is closed under multiplicative dependence if $n\in A$ implies every rational power of $n$ that is an integer $>1$ is also in $A$.

I've heard numbers that are normal to some bases but not normal to others be referred to as selectively normal.

As a curiosity, the problem of characterizing possible sets of simple normal bases was recently solved by Becher, Bugeaud, and Slaman (http://arxiv.org/pdf/1311.0332.pdf).

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    $\begingroup$ This answer solves the question which - I think - should have been asked to begin with. $\endgroup$ Sep 26, 2016 at 10:12

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